Math 1 Lecture 23. Dartmouth College. Wednesday
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1 Math 1 Lecture 23 Dartmouth College Wednesday
2 Contents Reminders/Announcements Last Time Derivatives of Logarithmic and Exponential Functions Examish Exercises
3 Reminders/Announcements WebWork due Friday Written HW due today x-hour tomorrow: proofs of inverse trig derivatives examish exercises
4 Last Time Implicit differentiation (f 1 ) (a) = 1/f (f 1 (a)) Derivatives of inverse trigonometric functions
5 dx (arcsin x) revisited
6 dx (arcsin x) revisited Let y = arcsin x. Then sin y = x and y [ π/2, π/2].
7 dx (arcsin x) revisited Let y = arcsin x. Then sin y = x and y [ π/2, π/2]. Using implicit differentiation (with respect to x) we get the equation (cos y) dy dx = 1.
8 dx (arcsin x) revisited Let y = arcsin x. Then sin y = x and y [ π/2, π/2]. Using implicit differentiation (with respect to x) we get the equation (cos y) dy dx = 1. Solving this equation for y we obtain dy dx = 1 cos y.
9 dx (arcsin x) revisited Let y = arcsin x. Then sin y = x and y [ π/2, π/2]. Using implicit differentiation (with respect to x) we get the equation (cos y) dy dx = 1. Solving this equation for y we obtain dy dx = 1 cos y. Now, since y [ π/2, π/2], we have cos y 0 and cos y = + 1 (sin y) 2 = 1 x 2.
10 dx (arcsin x) revisited Let y = arcsin x. Then sin y = x and y [ π/2, π/2]. Using implicit differentiation (with respect to x) we get the equation (cos y) dy dx = 1. Solving this equation for y we obtain dy dx = 1 cos y. Now, since y [ π/2, π/2], we have cos y 0 and cos y = + 1 (sin y) 2 = 1 x 2. Thus d dx (arcsin x) = 1 1 x 2.
11 dx (ln x)
12 dx (ln x) Let y = ln x so that e y = x.
13 dx (ln x) Let y = ln x so that e y = x. Then by differentiating this equation implicitly with respect to x we get e y dy dx = 1.
14 dx (ln x) Let y = ln x so that e y = x. Then by differentiating this equation implicitly with respect to x we get e y dy dx = 1. Solving this equation for dy dx yields dy dx = 1 e y = 1 x.
15 dx (ln x) Let y = ln x so that e y = x. Then by differentiating this equation implicitly with respect to x we get e y dy dx = 1. Solving this equation for dy dx yields dy dx = 1 e y = 1 x. The domain of this function is (0, ). Do you see why?
16 dx (ln x )
17 dx (ln x ) This is exactly the same as the previous example d dx (ln x ) = 1 x.
18 dx (ln x ) This is exactly the same as the previous example d dx (ln x ) = 1 x. The only difference is that this derivative is valid on (, 0) (0, ).
19 dx (ln x ) This is exactly the same as the previous example d dx (ln x ) = 1 x. The only difference is that this derivative is valid on (, 0) (0, ). Consider the graphs of ln(x) and ln x...
20 dx (log a x)
21 dx (log a x) First note that log a (x) = ln(x) ln(a) = 1 ln(x). ln(a) }{{} constant
22 dx (log a x) First note that Thus log a (x) = ln(x) ln(a) = 1 ln(x). ln(a) }{{} constant d dx (log a(x)) = d ( ) ln(x) dx ln(a) = 1 d ln(a) dx (ln(x)) = 1 ln(a) 1 x = 1 x ln(a).
23 dx (log a x )
24 dx (log a x ) We saw in the previous slide that d dx (log a(x)) = 1 x ln(a).
25 dx (log a x ) We saw in the previous slide that d dx (log a(x)) = 1 x ln(a). Just like the natural logarithm, this function (viewed as a derivative) is valid on the domain (0, ).
26 dx (log a x ) We saw in the previous slide that d dx (log a(x)) = 1 x ln(a). Just like the natural logarithm, this function (viewed as a derivative) is valid on the domain (0, ). We also have that d dx (log a x ) = 1 x ln(a).
27 dx (log a x ) We saw in the previous slide that d dx (log a(x)) = 1 x ln(a). Just like the natural logarithm, this function (viewed as a derivative) is valid on the domain (0, ). We also have that d dx (log a x ) = 1 x ln(a). What domain is this function valid on?
28 dx (log a x ) We saw in the previous slide that d dx (log a(x)) = 1 x ln(a). Just like the natural logarithm, this function (viewed as a derivative) is valid on the domain (0, ). We also have that d dx (log a x ) = 1 x ln(a). What domain is this function valid on? (, 0) (0, ).
29 dx (ax )
30 dx (ax ) The hint is to write a x = (e ln(a) ) x and use the chain rule.
31 dx (ax ) The hint is to write a x = (e ln(a) ) x and use the chain rule. d dx (ax ) = d ((e ) ln(a) ) x dx = d ) (ln a)x (e dx = e (ln a)x (ln a) = (ln a) a x.
32 dx (ax ) The hint is to write a x = (e ln(a) ) x and use the chain rule. If a = e does this check out? d dx (ax ) = d ((e ) ln(a) ) x dx = d ) (ln a)x (e dx = e (ln a)x (ln a) = (ln a) a x.
33 Logarithmic Differentiation
34 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations.
35 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations. We illustrate this by differentiating the function y = x 3/4 x 2 +1 (3x+2) 5.
36 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations. We illustrate this by differentiating the function y = x 3/4 x First we take the (3x+2) 5 natural logarithm of both sides of the equation and simplify to get ln(y) = 3 4 ln(x) ln(x 2 + 1) 5 ln(3x + 2).
37 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations. We illustrate this by differentiating the function y = x 3/4 x First we take the (3x+2) 5 natural logarithm of both sides of the equation and simplify to get ln(y) = 3 4 ln(x) ln(x 2 + 1) 5 ln(3x + 2). Now we differentiate implicitly (with respect to x) to get y = x + 1 2x 2 x x + 2. y
38 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations. We illustrate this by differentiating the function y = x 3/4 x First we take the (3x+2) 5 natural logarithm of both sides of the equation and simplify to get ln(y) = 3 4 ln(x) ln(x 2 + 1) 5 ln(3x + 2). Now we differentiate implicitly (with respect to x) to get y = x + 1 2x 2 x x + 2. y Lastly, solve the equation for y by clearing y from the denominator.
39 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations. We illustrate this by differentiating the function y = x 3/4 x First we take the (3x+2) 5 natural logarithm of both sides of the equation and simplify to get ln(y) = 3 4 ln(x) ln(x 2 + 1) 5 ln(3x + 2). Now we differentiate implicitly (with respect to x) to get y = x + 1 2x 2 x x + 2. y Lastly, solve the equation for y by clearing y from the denominator. Using logarithmic differentiation can turn incredibly annoying computations into simple applications of the chain rule.
40 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations. We illustrate this by differentiating the function y = x 3/4 x First we take the (3x+2) 5 natural logarithm of both sides of the equation and simplify to get ln(y) = 3 4 ln(x) ln(x 2 + 1) 5 ln(3x + 2). Now we differentiate implicitly (with respect to x) to get y = x + 1 2x 2 x x + 2. y Lastly, solve the equation for y by clearing y from the denominator. Using logarithmic differentiation can turn incredibly annoying computations into simple applications of the chain rule. It also allows us to compute derivatives of functions like y = x x...
41 Logarithmic Differentiation The technique of logaritmic differentiation can be used to dramatically simplify some derivative computations. We illustrate this by differentiating the function y = x 3/4 x First we take the (3x+2) 5 natural logarithm of both sides of the equation and simplify to get ln(y) = 3 4 ln(x) ln(x 2 + 1) 5 ln(3x + 2). Now we differentiate implicitly (with respect to x) to get y = x + 1 2x 2 x x + 2. y Lastly, solve the equation for y by clearing y from the denominator. Using logarithmic differentiation can turn incredibly annoying computations into simple applications of the chain rule. It also allows us to compute derivatives of functions like y = x x... you just win.
42 Examish Exercises Find the derivative of each function and the domain on which it is valid. 1. y = ln(x + 5) 2. y = ln x + 5
43 Examish Exercises 1. f (x) = x ln x x 2. f (x) = sin(ln x) 3. y = ln 1 x 4. g(x) = ln(xe 2x ) 5. f (x) = log 10 x 6. h(x) = log 10 x 7. y = 2 x 8. y = 5 2x+1 9. y = (x 2 + 2) 2 (x 4 + 4) y = (2x + 1) x
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